The product (A) formed in the following reaction sequence is : C H _ 3 - C ≡ C H [ (i i) H _ 2 / N i ](i) H g ^ 2 +, H _ 2 S O _ 4 (A)$$\mathrm {C H} _ {3} - \mathrm {C} \equiv \mathrm {C H} \xrightarrow [ (\mathrm {i i}) \mathrm {H} _ {2} / \mathrm {N i} ]{(\mathrm {i}) \mathrm {H} g ^ {2 +}, \mathrm {H} _ {2} \mathrm {S O} _ {4}} (A)$$

Related Formula

Phenolic protons react with standard strong bases:

Aldol condensation structural requirement: Requires presence of acidic α$\alpha$-hydrogen atoms connected to carbonyl centers.

Core Logic

Let's analyze functional groups within the Vanillin molecular framework:

• Vanillin contains a phenolic hydroxyl group, an aromatic ether, and a formyl functional group (benzaldehyde derivative).
• Statement I: The presence of the phenolic -OH$-\mathrm{OH}$ group allows acid-base reaction with NaOH$\mathrm{NaOH}$ directly

  

  . The aldehyde center readily reduces Tollen's reagent to produce a silver mirror. (Statement I is accurate).
• Statement II: Vanillin lacks any alpha-hydrogens adjacent to its carbonyl carbon, preventing it from undergoing self-aldol condensation. (Statement II is false).

Pattern Recognition

Benzaldehyde and its substituted derivatives (like vanillin or benzaldehyde itself) never undergo self-aldol condensation because they lack α$\alpha$-carbons with abstractable protons. Instead, they typically perform Cannizzaro transformations when exposed to highly concentrated alkaline media.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Related Formula

Intramolecular Cannizzaro-type reaction or Benzil-Benzilic acid rearrangement involves nucleophilic attack of hydroxide at a carbonyl group, followed by hydride transfer to the adjacent carbonyl carbon.

Core Logic

Let's analyze the starting compound, phenylglyoxal:

• The aldehyde carbon (-CHO$-CHO$) is much more electrophilic than the ketone carbon (-CO-$-CO-$) due to less steric hindrance and absence of phenyl group electron donation.
• Hydroxide ion (OH^-$\mathrm{OH}^-$) selectively attacks the aldehyde carbonyl carbon, forming a tetrahedral intermediate.

Step 1: Hydride Transfer Mechanism

The tetrahedral intermediate collapses, prompting an intramolecular hydride (H^-$H^-$) transfer to the adjacent ketone carbonyl carbon:

This is the rate-determining step.

Step 2: Proton Transfer to form Product

Rapid proton transfer occurs from the carboxylic acid group to the alkoxide oxygen, yielding the stable carboxylate salt:

This corresponds to Option (2).

Pattern Recognition

In asymmetrical 1,2-dicarbonyl systems with an aldehyde and a ketone, nucleophilic addition occurs preferentially at the more reactive aldehyde carbon. The hydrogen is then transferred as a hydride to the ketone carbon, yielding an α$\alpha$-hydroxy carboxylate salt.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Core Logic

Intramolecular aldol condensations are governed heavily by thermodynamic stability, favoring the formation of 5-membered or 6-membered rings over strained 3-, 4-, or large 7-membered options.

• Enolate Generation: Base abstracts an α$\alpha$-proton to form a nucleophilic carbanion enolate.
• Attack Vector Selection: Deprotonation at the outer methyl site enables a ring-closing attack on the distant carbonyl carbon, perfectly designing a highly stable cyclopentene ring skeleton.
• Dehydration: Heating drives the loss of a water molecule (-H₂O$-\text{H}_2\text{O}$), introducing an α,β$\alpha,\beta$-unsaturated carbonyl arrangement that provides stabilization via conjugated resonance.

  

Pattern Recognition

Count the intervening carbon chain carefully. Intramolecular aldol pathways will always selectively build 5- or 6-membered rings due to favorable ring strain kinetics. Eliminating choices based on incorrect ring sizes isolates Option (1) instantly.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Core Logic

Let us systematically follow the transformation steps:

• First Elimination: 1,2-dibromocyclooctane reacts with alcoholic KOH$\text{KOH}$ to remove one molecule of HBr$\text{HBr}$, resulting in a bromocyclooctene intermediate.
• Second Elimination: Treatment with the stronger base NaNH₂$\text{NaNH}_2$ removes the second molecule of HBr$\text{HBr}$, forming an alkyne inside the 8-membered ring: cyclooctyne.
• Kucherov Reaction: Hydration of cyclooctyne using Hg²⁺/H^+$\text{Hg}^{2+}/H^+$ creates an enol intermediate that undergoes tautomerization to form a stable ketone: cyclooctanone.
• Clemmensen Reduction: Subjecting cyclooctanone to zinc amalgam and hydrochloric acid (Zn-Hg/HCl$\text{Zn-Hg/HCl}$) completely reduces the carbonyl group (>C=O$>C=O$) to a methylene group (-CH₂-$-\text{CH}_2-$), finishing with cyclooctane.

  

Pattern Recognition

A vicinal dihalide treated with sequential strong bases creates an alkyne path. Alkyne hydration creates a ketone body. Finally, Clemmensen reduction takes the ketone down to a simple hydrocarbon skeleton. Recognizing this terminal reduction loop establishes cyclooctane as the undisputed answer.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Hydrocarbons

Core Logic

The balanced chemical equation for the synthesis of dibenzalacetone is:

Let's calculate the moles of reactants:

According to the reaction stoichiometry, 2 moles of PhCHO$2\text{ moles of } PhCHO$ yield 1 mole of dibenzalacetone$1\text{ mole of dibenzalacetone}$.

Step 1: Yield Evaluation

Pattern Recognition

Always remember the stoichiometric ratio: It takes 2 moles of benzaldehyde to condense with 1 mole of acetone to form the symmetrical dibenzalacetone product.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids